Antibody genetics and function

Question 1

What are the main molecules for antigen recognition?

Answer 1

PRRs, MHC, TCR and BCR. Antibodies which are glycoproteins, not main.

Question 2

Describe the structure of antibody molecules.

Answer 2

There are 2 identical heavy and 2 identical light chains which have disulphide bonds between light chains and heavy chains to bond them.

Question 3

What is the molecular weight of heavy chains?

Answer 3

50 kDa

Question 4

What is the molecular weight of light chains?

Answer 4

25 kDa

Question 5

What is the molecular weight of an antibody?

Answer 5

~150 kDa

Question 6

What enables antigen specificity of antibodies?

Answer 6

the variable regions of the antibody, which vary between one antibody and another. There are hypervariable regions on complementarity determining regions (CDR1/2/3). The amino acids at the CDR region are complementary to the amino acids on the epitope of the antigen.

Question 7

How many subclasses of IgG are there?

Answer 7

4

Question 8

How many subclasses of IgA, E, D and M?

Answer 8

Ig A has 2, however the others do not have subclasses.

Question 9

How many possible constant regions of antibodies are there?

Answer 9

9

Question 10

How does antibody variability arise?

Answer 10

By genetic recombination of Ig genes for the variability regions. They only recombine in B cells and have a germ line configuration

Question 11

Describe heavy chain genetics

Answer 11

They are syntenic and there are 4 groups of genes: Variable, Diversity, Joining and constant.

Question 12

How many variable genes are there?

Answer 12

~40

Question 13

How many diversity genes are there?

Answer 13

~27

Question 14

How many joining genes are there?

Answer 14

~6

Question 15

How many constant genes are there?

Answer 15

9

Question 16

Describe light chain genetics

Answer 16

They are not syntenic and do not have diversity genes, made up of 3 gene segments.

Question 17

Describe B cell development

Answer 17

They start in the bone marrow as haematopoietic stem cells and develop down the lymphoid pathway. B cells then recombine their Ig genes, T cells recombine TCR genes. For B cells, in germ line configuration, they recombine the VDJ gene segments in a semi-random way:
1/27 D genes are positioned next to 1/6 joining genes. Chromosomal inversion loops enable recombination.
2ndary recombination occurs, where 1/40 V segments are recombined with the DJ genes. Primary RNA transcript transcribed, introns removed from pre-mRNA (splicing). The mRNA then has VDJC mRNA. Recombination of kappa and lambda light chains also occurs; if kappa fails then lambda occurs.

Question 18

Which part of the antibody has the most variable structure?

Answer 18

The Ig heavy chains (1/40 x 1/27 x 1/6 = 1/6480)

Question 19

What is the variability of Igkappa?

Answer 19

40 x 5 = 200

Question 20

What is the variability of Iglambda?

Answer 20

30 x 5 = 150

Question 21

What is the Ig variability value?

Answer 21

200 (with kappa) x 6480 = 1.3 x 10^6

Question 22

How many kappa light chain genes are there?

Answer 22

50

Question 23

How many lambda light chain genes are there?

Answer 23

30

Question 24

How many constant genes are there for light chains?

Answer 24

5 for both kappa and lambda

Question 25

How is V-D-J recombination mediated?

Answer 25

By recombinase enzymes encoded by recombination-activating genes; RAG-1 and RAG-2 which recognise the recombination signal sequence (RSS) nucleotides flanking the V,D and J gene segments.

Question 26

Where do V segments have their RSS

Answer 26

3' downstream

Question 27

Where do D segments have their RSS?

Answer 27

at 5' and 3' flanking sites?

Question 28

Where do J segments have their RSS?

Answer 28

upstream 5'

Question 29

How many types of RSS are there?

Answer 29

2; heptamer (7) - nonamer (9) sequences.

Question 30

What is the sequence of RSS?

Answer 30

5'-heptamer-unconserved spacer - nonamer - 3' sequences.

Question 31

How long are unconserved spacers?

Answer 31

12 or 23 nucleotides long. The length and sequence of RSS is conserved at heptamer and nonamer sequences. This 12-23 rule means that a sequence can only be recombined that has an RSS of 12bp spacer to a sequence with a 23bp spacer. There is no recombiantion between V and a J not with 2 Ds. This 12-23 rule ensures that the sequence is V-D-J.

Question 32

How does additional diversity of antibodies arise?

Answer 32

recombinatorial inaccuracies, N-nucleotide addition using terminal deoxynucleotidyl transferase (TdT), chain combinations, somatic hypermutation

Question 33

What does class switching enable?

Answer 33

Allows expression of different antibody isotopes, IgM to IgE.

Question 34

Are RAG-1/2 used in the recombination of heavy chains?

Answer 34

No, different enzymes involved.

Question 35

Are there switch sequences between Cmu and Cdelta?

Answer 35

No, this results in IgM and IgD being constitutively expressed on naive B cells.

Question 36

How is TCR diversity generated?

Answer 36

Multiple germ line genes. VJ and VDJ recombination, recombinatorial inaccuracies, N-nucleotide addition and chain combinations.

Question 37

Can TCR diversity be generated through somatic hypermutation?

Answer 37

No, nor through class switch recombination.

Question 38

Describe antibody structure

Answer 38

Fab regions; the fragment of antigen binding region and an Fc region which can be split by papain. Fc region binds to FcR. Ab heavy and light chains are folded into domains. Domains are stabilised through intra-chain disulphide bonds. The hinge region enables flexibility of two Fab regions. The Ig domain structure becomes stable through folding. 4 beta strands make 2 beta sheets which are hydrophobic and intra=chain disulphide bonds stabilises. the hypervariable regions are exposed to the exterior. In the V domain, the antigen-binding CDRs are located in Ag-binding loops at the tip of the Ab.

Question 39

Can antibodies form polymers?

Answer 39

Yes, particularly secretory IgA; dimeric using J-chain protein. Circulating IgM can form a pentameric polymer with J chains linking.